If $f(x) = \frac{1+x}{1-x}$ where $x \neq 1$,then $f(x) \cdot f(y) = $ . . . . . . .

If the function $f(x) = \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$,then the domain of the composite function $(g \circ f)(x)$ is:

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

If $f(x)=3x+6$,$g(x)=4x+k$ and $f \circ g(x)=g \circ f(x)$,then $k =$

$f(x) = \begin{cases} 3-x, & -1 \leqslant x < 0 \\ 1+\frac{5x}{3}, & -3 \leqslant x \leqslant 2 \end{cases}$ and $g(x) = \begin{cases} -x, & -2 \leqslant x \leqslant 3 \\ x, & 0 \leqslant x \leqslant 1 \end{cases}$. Find the range of $(f \circ g)(x)$.

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$